Problem source

The function $\f$ satisfies $\f(x+1)= \f(x)$ and $\f(x)>0$ for all $x$. 
\begin{questionparts}
\item
Give an example of such a function.
\item
The function $\F$ satisfies
\[
\frac{\d \F}{\d x} =\f(x)
\]
and $\F(0)=0$.
Show that $\F(n) = n\F(1)$, for any positive integer $n$.
\item
Let $y$ be the solution of the differential equation
\[
\frac{\d y}{\d x} +\f(x) y=0
\]
that satisfies $y=1$ when $x=0$. Show that $y(n) \to 0$ as $n\to\infty$, where
$n= 1,\,2,\, 3,\, \ldots$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item $f(x) = \lfloor x \rfloor+1$

\item Clearly $\displaystyle F(x) = \int_0^x f(t) \d t$, in particular:
\begin{align*}
&& F(n) &= \int_0^n f(t) \d t \\
&&&= \sum_{i=1}^n \int_{i-1}^i f(t) \d t \\
&&&= \sum_{i=1}^n \int_{0}^1 f(t-i+1) \d t \\
&&&= \sum_{i=1}^n \int_{0}^1 f(t) \d t \\
&&&= n \int_{0}^1 f(t) \d t\\
&&&= n F(1)
\end{align*}

\item $\,$
\begin{align*}
&& 0 &= \frac{\d y}{\d x} +f(x) y \\
\Rightarrow && \int -f(x) \d x &= \int \frac1y \d y\\
\Rightarrow && -F(x) & = \ln y + C \\
x=0,y=1: && C &= -F(0) \\
\Rightarrow && y &= \exp(F(0)-F(x))
\end{align*}

Well this $F(0)-F(x)$ is equivalent to $-F(x)$ where $F(0) = 0$, in particular $F(n) = nF(1)$, so

$y(n) = e^{-nF(1)}$ which tends to zero as long as $F(1) > 0$, but since $f(x) > 0$ for all $x$ this must be true.

\end{questionparts}